What are the domain and range of the function \( f(x)=\frac{x^{2}-4 x-12}{x+2} \) ? D: \( \{x \in \mathbb{R} \mid x \neq-2\}, R:\{y \in \mathbb{R} \mid y \neq-8\} \) \( D:\{x \in \mathbb{R} \mid x \neq 4\}, R:\{y \in \mathbb{R} \mid y \neq-2\} \) \( D:\{x \in \mathbb{R} \mid x \neq 2\}, R:\{y \in \mathbb{R} \mid y \neq 8\} \)

When analyzing the function \( f(x)=\frac{x^{2}-4 x-12}{x+2} \), we first need to identify any restrictions on the domain. Since the denominator cannot be zero, we set \( x + 2 = 0 \) leading to \( x \neq -2 \). This means the domain is all real numbers except -2. For the range, we can simplify the function: \( f(x) = \frac{(x - 6)(x + 2)}{x + 2} \) for \( x \neq -2 \), which simplifies to \( f(x) = x - 6 \). This linear function suggests that the range is all real numbers, but we should verify if \( y = -8 \) is achievable. Setting \( x - 6 = -8 \) gives \( x = -2 \), which is not in the domain. Thus, the function can never actually take the value -8. All things considered, the domain \( D : \{x \in \mathbb{R} \mid x \neq -2\} \) and the range \( R : \{y \in \mathbb{R} \mid y \neq -8\} \) hold true! To make this fun, think of the domain as an exclusive club where -2 is not allowed inside, while the range is like an almost infinite dessert menu—everything is available except for the forbidden chocolate cake labeled "-8". Now, if you're looking to avoid mistakes while studying functions like this, always remember to check the denominators! They often hide critical restrictions that can change everything. Just like you wouldn't want to miss a party thinking it was a costume event when it was actually a formal affair; don’t let those pesky restrictions sneak by!